View source for Every map from a space with the discrete topology is continuous

==Statement==
Given two [[Topological space|topologies]], {{M|(X,\mathcal{P}(X))}} and {{M|(Y,\mathcal{J})}} where:
* {{M|(X,\mathcal{P}(X))}} denotes the [[Discrete topology|discrete topology on {{M|X}}]]
We have the following<ref name="ALEC">Alec's own work</ref>:
* Any [[Function|mapping]], {{M|1=f:X\rightarrow Y}}, is [[Continuous map|continuous]]

==Proof of claim==
Recall there are two definitions of continuity, the ''topological'':
* {{M|1=f:(X,\mathcal{J})\rightarrow(Y,\mathcal{K})}} is continuous if {{M|1=\forall A\in\mathcal{K}[f^{-1}(A)\in\mathcal{J}]}} - for all open sets in {{M|Y}} the pre-image under {{M|f}} is open in {{M|X}}
Then there's the ''[[Metric space|metric space]]'' definition:
* {{M|1=f:(X,d_1)\rightarrow(Y,d_2)}} is continuous if {{M|1=\forall x\in X\forall\epsilon\in\mathbb{R}_{>0}\exists\delta\in\mathbb{R}_{>0}\forall y\in X[d_1(x,y)<\delta\implies d_1(f(x),f(y))<\epsilon]}}
===Topological sense of continuity===
'''Proof:'''
* We wish to prove: {{M|1=\forall A\in\mathcal{J}[f^{-1}(A)\in\mathcal{P}(X)]}}
: Let {{M|A\in\mathcal{J} }} be given. 
:: Then {{M|f^{-1}(A)\subseteq X}} by definition of pre-image
::* Which recall is: {{M|1=f^{-1}(A):=\{x\in X\vert f(x)\in A\} }}
:: As every subset of {{M|X}} is [[Open set|open]] in {{M|X}} we see that {{M|f^{-1}(A)}} is open.
: As {{M|A}} was arbitrary, this completes the proof.
* This isn't surprising as by definition {{M|A\subseteq X\iff A\in\mathcal{P}(X)}} so we really wanted to show {{M|1=\forall A\in\mathcal{J}[f^{-1}(A)\subseteq X]}}
There is no need to proceed and consider the metric space, however it is included for completeness. 
===Metric sense of continuity===
Suppose that {{M|(X,d)}} is a [[Metric space|metric space]] where {{M|d}} denotes the [[Discrete metric|discrete metric]] and {{M|(Y,d')}} is any metric space.<br/>We wish to show:
* {{M|1=\forall x\in X\forall\epsilon\in\mathbb{R}_{>0}\exists\delta\in\mathbb{R}_{>0}\forall y\in X[d(x,y)<\delta\implies d'(f(x),f(y))<\epsilon]}}
'''Proof:'''
: Let {{M|x\in X}} be given
:: Let {{M|\epsilon > 0}} be given
::: Choose {{M|1=\delta=\tfrac{1}{2} }}
:::: Let {{M|y\in X}} be given
::::: If {{M|1=d(x,y)<\tfrac{1}{2} }} we must have {{M|1=x=y}}
:::::* (As {{M|1=x\ne y\implies d(x,y)=1}} so by [[Contrapositive|contrapositive]] we have {{M|1=d(x,y)\ne 1\implies x=y}})
:::: As {{M|1=x=y}} we see {{M|1=d(x,y)=0}} and {{M|0<\delta}}
:::: y definition of a function {{M|1=x=y\implies f(x)=f(y)}} (a function must map a point to exactly one point of the image - it cannot map to two things)
:::: By definition of a metric {{M|1=f(x)=f(y)\implies d'(f(x),f(y))=0}}
:: {{M|1=d'(f(x),f(y))=0<\epsilon}} so {{M|d'(f(x),f(y))<\epsilon}}
: As {{M|x}} was arbitrary we have shown this is true for all {{M|x\in X}}
That completes the proof.

==References==
<references/>
{{Theorem Of|Metric Space|Topology}}
[[Category:First-year friendly]]
[[Category:Exemplary page]]